Efficiently Solving a Cubic Equation Using Synthetic Division

[FaraDazed]

Homework Statement

Find the real solutions of x^3+6x^2-8x+1

Homework Equations

quadratic

The Attempt at a Solution

First I tried simple numbers starting with 1 and found that +1 was a solution and therefore I could write

<br /> x^3+6x^2-8x+1=(x-1)(Ax^2+Bx+C) \\<br /> x^3+6x^2-8x+1=Ax^3+Bx^2+Cx-Ax^2-Bx-C \\<br />

And then equated the coeffeciants

For x^3, 1=A
For x^2, 6=B-A
For x^1, -8=C-B
For x^0, 1=-C

With knowing that A=1 and C= -1 , B can be found to be 7 .

And therefore I can write x^3+6x^2-8x+1=(x-1)(x^2+7x-1)

If I then use the quadratic formula on the quadratic factor I end up with
\frac{-7\pm\sqrt{53}}{2}

And as 53 is prime that is as far as it can go without losing accuracy.

I don't know if this is all I have to do, if I have done what I have done correctly or what. I am a bit confused. Any help/advice appreciated.

 

[Office_Shredder]

That looks entirely correct, and is all you have to do. You can include a line at the end that says

Therefore, the solutions are x=1,\ x=\frac{-7+\sqrt{53}}{2} and x=\frac{-7-\sqrt{53}}{2}

 

[LCKurtz]

Homework Statement

Find the real solutions of x^3+6x^2-8x+1

Homework Equations

quadratic

The Attempt at a Solution

First I tried simple numbers starting with 1 and found that +1 was a solution and therefore I could write

<br /> x^3+6x^2-8x+1=(x-1)(Ax^2+Bx+C) \\<br /> x^3+6x^2-8x+1=Ax^3+Bx^2+Cx-Ax^2-Bx-C \\<br />

And then equated the coeffeciants

For x^3, 1=A
For x^2, 6=B-A
For x^1, -8=C-B
For x^0, 1=-C

With knowing that A=1 and C= -1 , B can be found to be 7 .

And therefore I can write x^3+6x^2-8x+1=(x-1)(x^2+7x-1)

I'm curious why you didn't just divide the cubic by $x-1$ using either long division or, preferably, synthetic division to get the quadratic factor. Much less work...